Light as Particles

Planck quantized the energy exchanged between matter and radiation (P.2.1); Einstein, in 1905, quantized the light itself. This lesson follows the evidence that forced the photon on a reluctant physics community: the photoelectric effect, X-ray production, Compton scattering, and the creation and annihilation of matter. By the end, light — which Young's fringes had settled as a wave — is also, undeniably, a hail of particles. Light, like a cat, declines to be only one thing.

Learning Objectives

After this lesson you will be able to:

1. List the experimental facts of the photoelectric effect and explain, point by point, why the classical wave picture fails on each.
2. Apply Einstein's photon hypothesis $E = h\nu$ to compute work functions, $K_{max}$, and stopping potentials, and explain what Millikan's slope $h/e$ measures.
3. Derive the photon momentum $p = E/c = h/\lambda$ from the relativistic energy–momentum relation.
4. Derive the Duane–Hunt cutoff $\lambda_{min} = hc/eV$ and recognize it as the photoelectric effect in reverse.
5. Derive the Compton shift $\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta)$ from relativistic energy and momentum conservation.
6. Explain the threshold and kinematics of pair production and annihilation, including why each requires (or produces) more than one body.

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Intuition

Classically, light is a continuous electromagnetic wave: energy flux proportional to the squared amplitude (intensity), spread smoothly over the whole wavefront. Einstein's proposal replaces the smooth flood with rain: light of frequency $\nu$ arrives in discrete packets — photons — each of energy $E = h\nu$. Brightness counts how many packets arrive per second; color sets how much energy each carries; one photon is absorbed by one electron, whole or not at all. Every "impossible" photoelectric fact below becomes obvious in this picture — and the same bookkeeping, pushed through special relativity, predicts the Compton shift to four significant figures.

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Theory

The photoelectric effect: the facts, and why the wave picture fails on each

Shine light on a clean metal in vacuum; electrons come off and are collected, giving a photocurrent. Four facts, established between Hertz (1887) and Millikan (1916) — each paired, point by point, with what a classical wave (energy delivered continuously, at a rate $\propto E_0^2$, spread over the wavefront) predicts:

1. Threshold frequency. Each metal has a $\nu_0$ below which no electrons are emitted, at any intensity. Classically, any frequency should eject electrons if the light is bright enough — there should be no threshold.
2. No time lag. Emission starts within $<10^{-9}\,\mathrm{s}$, even in feeble light. Classically, an atom of radius $\sim 0.1\,\mathrm{nm}$ intercepts so little of a weak wavefront that accumulating a few eV should take months (E3: 135 days).
3. Intensity controls the current, not the electron energy. Doubling intensity doubles the photocurrent; $K_{max}$ is unchanged. Classically, a larger amplitude shakes electrons harder, so $K_{max}$ should grow with intensity.
4. Stopping potential is linear in frequency. The reverse voltage $V_0$ that just stops the fastest electrons ($eV_0 = K_{max}$) grows linearly with $\nu$, with the same slope for every metal. Classically, frequency is nearly irrelevant to energy transfer — no universal slope, let alone one equal to $h/e$ with the blackbody $h$ of P.2.1.

Einstein's photon and the photoelectric equation

Einstein (1905): light of frequency $\nu$ consists of quanta of energy $E = h\nu$, with $h = 6.626\times10^{-34}\,\mathrm{J\,s} = 4.136\times10^{-15}\,\mathrm{eV\,s}$. One photon is absorbed by one electron, which pays the work function $\phi$ — its minimum binding energy to the metal, a few eV — and leaves with at most

$$K_{max} = h\nu - \phi, \qquad\text{so}\qquad eV_0 = h\nu - \phi \;\Longrightarrow\; V_0 = \frac{h}{e}\,\nu - \frac{\phi}{e}.$$

Every fact follows. Threshold: $\nu > \nu_0 = \phi/h$ — below it no single photon suffices (and two-photon absorption is fantastically unlikely at ordinary intensities). No lag: the energy arrives in one lump. Intensity: more photons per second means more electrons per second, but each still gets exactly $h\nu$. And $V_0(\nu)$ is a straight line of metal-independent slope $h/e$, intercept $-\phi/e$. Millikan spent a decade trying to refute this and in 1916 confirmed it, measuring $h = 6.57\times10^{-34}\,\mathrm{J\,s}$ — within 0.5% of the blackbody value. Independent phenomena, same constant.

Photon momentum

Special relativity relates energy, momentum, and mass by $E^2 = (pc)^2 + (mc^2)^2$. A photon travels at $c$, which no massive particle can do at finite energy, so $m = 0$ and

$$E = pc \quad\Longrightarrow\quad p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda},$$

using $c = \lambda\nu$. The momentum is tiny — a green photon carries $1.2\times10^{-27}\,\mathrm{kg\,m/s}$ (E2) — but real: it reproduces Maxwell's radiation pressure $p = E/c$ in bulk, and Compton verified it photon by photon.

X-rays: the photoelectric effect in reverse

X-rays are made by accelerating electrons through $V \sim 10$–$100\,\mathrm{kV}$ into a metal anode. Decelerating in the nuclear electric fields, they radiate a continuous spectrum — bremsstrahlung ("braking radiation") — plus sharp characteristic lines from shell transitions (P.2.2). Classically the continuum should extend to arbitrarily short wavelength; instead it stops dead at the Duane–Hunt cutoff. The photon explains it in one line: an electron arrives with kinetic energy $eV$, and the most energetic photon takes all of it in one quantum,

$$h\nu_{max} = eV \quad\Longrightarrow\quad \lambda_{min} = \frac{c}{\nu_{max}} = \frac{hc}{eV}.$$

This is the photoelectric effect run backwards — electron in, photon out ($\phi$, a few eV against tens of keV, is negligible). The cutoff depends only on $V$, never on the anode material, and historically gave one of the cleanest measurements of $h$; with $hc = 1240\,\mathrm{eV\,nm}$, $\lambda_{min}[\mathrm{nm}] = 1.240/V[\mathrm{kV}]$, e.g. $31.0\,\mathrm{pm}$ at $40\,\mathrm{kV}$. Meanwhile the same X-rays reflect off crystal planes with constructive interference at the Bragg condition $2d\sin\theta = n\lambda$ — pure wave behavior, and the standard way X-ray wavelengths are measured. Hold that thought.

The Compton effect

Compton (1923) scattered $\lambda = 71.1\,\mathrm{pm}$ X-rays off graphite and found scattered radiation at angle $\theta$ with a longer wavelength $\lambda'$, shifted by an amount depending only on $\theta$. A classical wave cannot do this — an electron driven at $\nu$ reradiates at $\nu$ — but a relativistic photon–electron collision can. Setup: a photon ($E_\gamma = hc/\lambda$, momentum $h/\lambda$ along $x$) hits a free electron at rest; the photon exits at $\theta$ with wavelength $\lambda'$, the electron at $\varphi$ on the other side with momentum $p_e$ and energy $E_e = \sqrt{p_e^2c^2 + m_e^2c^4}$. Energy, $x$- and $y$-momentum conservation give three equations:

$$\frac{hc}{\lambda} + m_ec^2 = \frac{hc}{\lambda'} + \sqrt{p_e^2c^2 + m_e^2c^4}, \qquad \frac{h}{\lambda} = \frac{h}{\lambda'}\cos\theta + p_e\cos\varphi, \qquad 0 = \frac{h}{\lambda'}\sin\theta - p_e\sin\varphi .$$

Step 1 — eliminate $\varphi$. Isolate the electron terms in the momentum pair, square, add ($\cos^2\varphi + \sin^2\varphi = 1$):

$$p_e^2 = \Big(\frac{h}{\lambda}\Big)^2 + \Big(\frac{h}{\lambda'}\Big)^2 - \frac{2h^2}{\lambda\lambda'}\cos\theta .$$

Step 2 — eliminate $E_e$. Isolate the square root in the energy equation and square:

$$p_e^2c^2 = \Big(\frac{hc}{\lambda} - \frac{hc}{\lambda'} + m_ec^2\Big)^2 - m_e^2c^4 = \Big(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\Big)^2 + 2m_ec^2\Big(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\Big).$$

Step 3 — equate. Multiply Step 1 by $c^2$ and set it equal to Step 2; the squared terms $(hc/\lambda)^2 + (hc/\lambda')^2$ cancel, leaving $-\tfrac{2(hc)^2}{\lambda\lambda'}\cos\theta = -\tfrac{2(hc)^2}{\lambda\lambda'} + 2m_ec^2\,hc\,\tfrac{\lambda'-\lambda}{\lambda\lambda'}$. Multiply by $\lambda\lambda'/(2hc\,m_ec^2)$:

$$\boxed{\;\Delta\lambda = \lambda' - \lambda = \frac{h}{m_ec}\,(1-\cos\theta)\;}$$

The prefactor is the electron's Compton wavelength $\lambda_C = h/m_ec = 2.426\,\mathrm{pm}$, so the shift runs from $0$ (forward) to $2\lambda_C = 4.85\,\mathrm{pm}$ (backscatter), independent of $\lambda$ and of the material — exactly what Compton measured, with the recoil electron detected in coincidence with the scattered photon. In energy terms, with $\varepsilon = E_\gamma/m_ec^2$, the scattered photon has $E'_\gamma = E_\gamma/[1 + \varepsilon(1-\cos\theta)]$.

The unshifted peak. Compton also saw a component at the original $\lambda$: photons scattering off electrons so tightly bound that the whole atom recoils. Replace $m_e$ by the atomic mass $M \sim 10^4\,m_e$ and the shift $h/Mc$ shrinks by the same factor — unresolvably small.

Caution. Do not swap one cartoon for another: the photon is not a tiny billiard ball. The very same X-ray beam Bragg-diffracts off a crystal (wave) and Compton-scatters off its electrons (particle). No classical object does both; the resolution is that a quantum amplitude propagates like a wave and is detected like a particle — the formalism of P.3.2 and P.4.1.

Pair production and annihilation

Above roughly 1 MeV a new channel opens: the photon vanishes and an electron–positron pair appears, $\gamma \to e^- + e^+$, with threshold set by the rest masses, $E_\gamma \ge 2m_ec^2 = 1.022\,\mathrm{MeV}$. But this cannot happen in free space: each lepton has $E_i = \sqrt{p_i^2c^2 + m_e^2c^4} > p_ic$ strictly (because $m_e > 0$), so by the triangle inequality

$$E_- + E_+ > (p_- + p_+)\,c \ge |\vec p_- + \vec p_+|\,c ,$$

i.e. the pair always has $E > pc$, while the photon has exactly $E = pc$ — energy and momentum cannot both balance. Pair production therefore happens only near a nucleus, which absorbs the recoil momentum while (being heavy, $\Delta E = \Delta p^2/2M$) carrying off negligible energy, leaving the threshold at $1.022\,\mathrm{MeV}$. The time-reverse is annihilation: a slowed positron meets an electron essentially at rest, total momentum $\approx 0$; a single photon would carry $p = E/c \ne 0$ — forbidden by the same argument — so annihilation at rest produces at least two photons, back-to-back with equal and opposite momenta, $E_\gamma = m_ec^2 = 511\,\mathrm{keV}$ each (three-photon decay also occurs, from triplet positronium; two is the minimum). These back-to-back photon pairs are the working principle of PET imaging.

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Worked Examples

Example 1 — Photoelectric effect on sodium

Sodium has $\phi = 2.28\,\mathrm{eV}$; shine $\lambda = 400\,\mathrm{nm}$ light on it. Photon energy (with $hc = 1240\,\mathrm{eV\,nm}$): $E = hc/\lambda = 1240/400 = 3.10\,\mathrm{eV}$, so $K_{max} = 3.10 - 2.28 = 0.82\,\mathrm{eV}$ and $V_0 = 0.82\,\mathrm{V}$. Threshold: $\nu_0 = \phi/h = 2.28/(4.136\times10^{-15}) = 5.51\times10^{14}\,\mathrm{Hz}$, i.e. $\lambda_0 = 1240/2.28 = 544\,\mathrm{nm}$ — green barely works, red does nothing. Doubling the intensity at 400 nm doubles the current; $K_{max}$ and $V_0$ do not move.

Example 2 — Compton scattering at 90°

Compton's beam, $\lambda = 71.1\,\mathrm{pm}$ (Mo K$\alpha$), scattered at $\theta = 90^\circ$: $\Delta\lambda = \lambda_C(1 - \cos 90^\circ) = 2.43\,\mathrm{pm}$, so $\lambda' = 73.5\,\mathrm{pm}$. Energies: $E_\gamma = 1240/0.0711 = 17.4\,\mathrm{keV}$, $E'_\gamma = 1240/0.0735 = 16.9\,\mathrm{keV}$; cross-check with $\varepsilon = 17.4/511 = 0.0341$: $E' = 17.4/1.0341 = 16.9\,\mathrm{keV}$. ✓ The recoil electron carries $K_e \approx 0.58\,\mathrm{keV}$ — small here, but MeV photons can hand the electron most of their energy (E5).

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Hands-on (Python)

Fit synthetic stopping-potential data the way Millikan did — the slope of $V_0(\nu)$ is $h/e$ for every metal; only the intercept $-\phi/e$ changes.

import numpy as np

h_true = 6.62607015e-34      # Planck constant, J s
e      = 1.602176634e-19     # elementary charge, C
rng    = np.random.default_rng(7)
metals = {"Cs": 2.14, "Na": 2.28, "Zn": 4.33}    # work functions phi, eV

for metal, phi in metals.items():
    nu0 = phi * e / h_true                        # threshold frequency, Hz
    nu  = np.linspace(1.05 * nu0, 2.5 * nu0, 12)  # measured frequencies, Hz
    V0  = (h_true / e) * nu - phi                 # ideal line: slope h/e, intercept -phi/e
    V0 += rng.normal(0.0, 0.02, V0.size)          # 20 mV of measurement noise
    slope, intercept = np.polyfit(nu, V0, 1)      # least-squares straight line
    print(f"{metal}: h = {slope * e:.3e} J s   phi = {-intercept:.2f} eV")

# Expected output — h within ~1% of 6.626e-34 for every metal, e.g.:
# Cs: h = 6.645e-34 J s   phi = 2.15 eV   (Na, Zn similar; varies with the noise)

Now the Compton kinematics: the wavelength shift is universal, but the energy lost depends strongly on the incident energy.

import numpy as np
import matplotlib.pyplot as plt

h, c, me = 6.62607015e-34, 2.99792458e8, 9.1093837015e-31
lam_C    = h / (me * c)                          # Compton wavelength = 2.426e-12 m
mec2     = 0.51099895                            # electron rest energy, MeV
theta    = np.linspace(0.0, np.pi, 500)

fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10, 4))
ax1.plot(np.degrees(theta), lam_C * (1 - np.cos(theta)) * 1e12)
ax1.set(xlabel="θ (deg)", ylabel="Δλ (pm)", title="Compton shift")
# rises from 0 to 2*lam_C = 4.85 pm at 180°, independent of incident energy

for E in (0.1, 0.5, 1.0):                        # incident photon energies, MeV
    ax2.plot(np.degrees(theta), E / (1 + (E / mec2) * (1 - np.cos(theta))), label=f"E = {E} MeV")
ax2.set(xlabel="θ (deg)", ylabel="E' (MeV)", title="scattered photon energy")
ax2.legend()
plt.tight_layout()
plt.show()
# At θ = 180° the 1 MeV curve drops to 1/(1 + 2·1.957) ≈ 0.20 MeV:
# harder photons lose a larger FRACTION of their energy to the electron.

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Exercises

E1 (easy). Cesium has $\phi = 2.14\,\mathrm{eV}$. Can a helium–neon laser ($\lambda = 632.8\,\mathrm{nm}$) eject photoelectrons from it at any intensity? Give the threshold wavelength.

Solution

$E = 1240/632.8 = 1.96\,\mathrm{eV} < 2.14\,\mathrm{eV}$: no single photon suffices, so no emission at any (ordinary) intensity — intensity only changes the photon rate. Threshold: $\lambda_0 = hc/\phi = 1240/2.14 = 579\,\mathrm{nm}$; anything redder fails.

E2 (easy). A 1.00 mW green laser pointer ($\lambda = 532\,\mathrm{nm}$) shines on a black wall. Find the photon energy, the photons emitted per second, the momentum per photon, and the force on the wall.

Solution

$E = 1240/532 = 2.33\,\mathrm{eV} = 3.73\times10^{-19}\,\mathrm{J}$; rate $N = P/E = 10^{-3}/3.73\times10^{-19} = 2.7\times10^{15}\,\mathrm{s^{-1}}$; momentum $p = h/\lambda = 1.25\times10^{-27}\,\mathrm{kg\,m/s}$; force (full absorption) $F = Np = 3.3\times10^{-12}\,\mathrm{N}$ — equal to the classical $P/c$, as it must be.

E3 (medium). Classical time-lag estimate: sodium ($\phi = 2.28\,\mathrm{eV}$) is illuminated at $I = 1.0\,\mathrm{\mu W/m^2}$. If an atom of radius $r = 0.10\,\mathrm{nm}$ could only absorb the wave energy falling on its cross-section, how long until one electron accumulates $\phi$? Compare with experiment.

Solution

Collected power $P = I\pi r^2 = 10^{-6}\pi(10^{-10})^2 = 3.1\times10^{-26}\,\mathrm{W}$; required energy $\phi = 3.65\times10^{-19}\,\mathrm{J}$; time $t = 3.65\times10^{-19}/3.1\times10^{-26} = 1.2\times10^{7},\mathrm{s} \approx 135$ days. Observed lag: under a nanosecond. The energy is not spread over the wavefront — it arrives in quanta.

E4 (medium). Why did Compton need X-rays? Compute the maximum fractional shift $\Delta\lambda/\lambda$ for (a) green light, $\lambda = 500\,\mathrm{nm}$, and (b) Compton's X-rays, $\lambda = 71.1\,\mathrm{pm}$.

Solution

The maximum shift ($\theta = 180^\circ$) is $2\lambda_C = 4.85\,\mathrm{pm}$ regardless of $\lambda$. (a) $4.85\,\mathrm{pm}/500\,\mathrm{nm} \approx 10^{-5}$ — buried below optical resolution and linewidth. (b) $4.85/71.1 = 6.8\%$ — easily resolved by Bragg spectrometry. The absolute shift is universal; only short wavelengths make it relatively visible.

E5 (hard). Derive the maximum kinetic energy the electron can gain in Compton scattering (the Compton edge), and evaluate it for a 511 keV photon.

Solution

The electron gains most when the photon backscatters, $\theta = 180^\circ$: with $\varepsilon = E_\gamma/m_ec^2$, $E'_\gamma = E_\gamma/(1+2\varepsilon)$, so $K_{max} = E_\gamma - E'_\gamma = E_\gamma\,\frac{2\varepsilon}{1+2\varepsilon}$. For $E_\gamma = 511\,\mathrm{keV}$, $\varepsilon = 1$: $K_{max} = 511\times\tfrac23 = 341\,\mathrm{keV}$. This sharp upper edge in the electron spectrum is a standard feature of gamma-ray detectors — the photon can never give a free electron all its energy (total absorption would violate the same $E$-vs-$pc$ bookkeeping as pair production in free space).

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Checkpoint

1. Which photoelectric facts contradict the classical wave picture, and how does $E = h\nu$ explain each?
2. What does the slope of Millikan's $V_0$-vs-$\nu$ line measure, and why is it the same for all metals?
3. Derive $p = h/\lambda$ for the photon. Which relativistic ingredient makes it work?
4. Why does Compton-scattered radiation contain an unshifted line as well as the shifted one?
5. Why is pair production impossible in empty space, and why does annihilation at rest yield two photons, not one?

Answers

1. Threshold frequency, no time lag, intensity-independence of $K_{max}$, linear $V_0(\nu)$. Photons: one quantum $h\nu$ per electron — below $\phi$ nothing works, energy arrives in a lump, intensity only sets the photon rate, and $eV_0 = h\nu - \phi$ is linear by construction.
2. The universal slope $h/e$ — nature's constants only; the metal enters solely via the intercept $-\phi/e$.
3. From $E^2 = (pc)^2 + (mc^2)^2$ with $m = 0$: $E = pc$, so $p = E/c = h\nu/c = h/\lambda$. The ingredient: a particle moving at $c$ must be massless.
4. Some photons scatter off tightly bound electrons — effectively off the whole atom — so the shift $h/Mc$ with $M \sim 10^4\,m_e$ is unresolvably small.
5. A pair always has $E > pc$ while a photon has $E = pc$, so both conservation laws cannot hold without a nucleus taking recoil momentum. At rest the total momentum is zero, which one photon ($p = E/c \ne 0$) cannot match — two back-to-back 511 keV photons can.

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Further Reading

• [ER] Eisberg & Resnick, §2-2–2-4 — photoelectric effect, Einstein's quantum theory, and the Compton effect; the primary treatment this lesson follows.
• [ER] Eisberg & Resnick, §2-6–2-7 — X-ray production, pair production and annihilation.
• [Gri] Griffiths & Schroeter, §1.1–1.2 — where the story hands over to the wavefunction and the statistical interpretation.

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